which satisfies the boundary conditions \(F(0)=1\)  and
\(F^{\prime }(0)=0\)  near the boundary \(z=0\)  [1]}. Inserting
Eq. (REF ) into Eq. (REF ), one can show that the
trial function satisfies the following second-order SL self
adjoint differential equation
\((1-z^3)F^{\prime \prime }(z)-3z^2F^{\prime }(z)-zF(z)+\frac{\lambda ^2(1-z)}{(1+z+z^2)^2}\left(1-\frac{4b\lambda ^2}{5}\zeta (z)\right)F(z)=0.\) 
